1.07t Construct differential equations: in context

10 A model for the height, \(h\) metres, of a certain type of tree at time \(t\) years after being planted assumes that, while the tree is growing, the rate of increase in height is proportional to \(( 9 - h ) ^ { \frac { 1 } { 3 } }\). It is given that, when \(t = 0 , h = 1\) and \(\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.2\).

1. Show that \(h\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.1 ( 9 - h ) ^ { \frac { 1 } { 3 } } .$$
2. Solve this differential equation, and obtain an expression for \(h\) in terms of \(t\).
3. Find the maximum height of the tree and the time taken to reach this height after planting.
4. Calculate the time taken to reach half the maximum height.

6 A certain curve is such that its gradient at a point \(( x , y )\) is proportional to \(x y\). At the point \(( 1,2 )\) the gradient is 4 .

1. By setting up and solving a differential equation, show that the equation of the curve is \(y = 2 \mathrm { e } ^ { x ^ { 2 } - 1 }\).
2. State the gradient of the curve at the point \(( - 1,2 )\) and sketch the curve.

7 \includegraphics[max width=\textwidth, alt={}, center]{e26f21c5-3776-4c86-8440-6959c5e37486-12_444_382_258_886} A water tank has vertical sides and a horizontal rectangular base, as shown in the diagram. The area of the base is \(2 \mathrm {~m} ^ { 2 }\). At time \(t = 0\) the tank is empty and water begins to flow into it at a rate of \(1 \mathrm {~m} ^ { 3 }\) per hour. At the same time water begins to flow out from the base at a rate of \(0.2 \sqrt { } h \mathrm {~m} ^ { 3 }\) per hour, where \(h \mathrm {~m}\) is the depth of water in the tank at time \(t\) hours.

1. Form a differential equation satisfied by \(h\) and \(t\), and show that the time \(T\) hours taken for the depth of water to reach 4 m is given by $$T = \int _ { 0 } ^ { 4 } \frac { 10 } { 5 - \sqrt { } h } \mathrm {~d} h$$
2. Using the substitution \(u = 5 - \sqrt { } h\), find the value of \(T\).

9 In an experiment to study the spread of a soil disease, an area of \(10 \mathrm {~m} ^ { 2 }\) of soil was exposed to infection. In a simple model, it is assumed that the infected area grows at a rate which is proportional to the product of the infected area and the uninfected area. Initially, \(5 \mathrm {~m} ^ { 2 }\) was infected and the rate of growth of the infected area was \(0.1 \mathrm {~m} ^ { 2 }\) per day. At time \(t\) days after the start of the experiment, an area \(a \mathrm {~m} ^ { 2 }\) is infected and an area \(( 10 - a ) \mathrm { m } ^ { 2 }\) is uninfected.

1. Show that \(\frac { \mathrm { d } a } { \mathrm {~d} t } = 0.004 a ( 10 - a )\).
2. By first expressing \(\frac { 1 } { a ( 10 - a ) }\) in partial fractions, solve this differential equation, obtaining an expression for \(t\) in terms of \(a\).
3. Find the time taken for \(90 \%\) of the soil area to become infected, according to this model.

8 In a certain chemical reaction the amount, \(x\) grams, of a substance present is decreasing. The rate of decrease of \(x\) is proportional to the product of \(x\) and the time, \(t\) seconds, since the start of the reaction. Thus \(x\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - k x t$$ where \(k\) is a positive constant. At the start of the reaction, when \(t = 0 , x = 100\).

1. Solve this differential equation, obtaining a relation between \(x , k\) and \(t\).
2. 20 seconds after the start of the reaction the amount of substance present is 90 grams. Find the time after the start of the reaction at which the amount of substance present is 50 grams.

6

1. By sketching a suitable pair of graphs, show that the equation $$2 - x = \ln x$$ has only one root.
2. Verify by calculation that this root lies between 1.4 and 1.7.
3. Show that this root also satisfies the equation $$x = \frac { 1 } { 3 } ( 4 + x - 2 \ln x )$$
4. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 3 } \left( 4 + x _ { n } - 2 \ln x _ { n } \right)$$ with initial value \(x _ { 1 } = 1.5\), to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

8 \includegraphics[max width=\textwidth, alt={}, center]{c687888e-bef0-4ea9-b5b3-e614028cc07c-3_654_805_274_671} An underground storage tank is being filled with liquid as shown in the diagram. Initially the tank is empty. At time \(t\) hours after filling begins, the volume of liquid is \(V \mathrm {~m} ^ { 3 }\) and the depth of liquid is \(h \mathrm {~m}\). It is given that \(V = \frac { 4 } { 3 } h ^ { 3 }\). The liquid is poured in at a rate of \(20 \mathrm {~m} ^ { 3 }\) per hour, but owing to leakage, liquid is lost at a rate proportional to \(h ^ { 2 }\). When \(h = 1 , \frac { \mathrm {~d} h } { \mathrm {~d} t } = 4.95\).

1. Show that \(h\) satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { 5 } { h ^ { 2 } } - \frac { 1 } { 20 } .$$
2. Verify that \(\frac { 20 h ^ { 2 } } { 100 - h ^ { 2 } } \equiv - 20 + \frac { 2000 } { ( 10 - h ) ( 10 + h ) }\).
3. Hence solve the differential equation in part (i), obtaining an expression for \(t\) in terms of \(h\).

10 In a model of the expansion of a sphere of radius \(r \mathrm {~cm}\), it is assumed that, at time \(t\) seconds after the start, the rate of increase of the surface area of the sphere is proportional to its volume. When \(t = 0\), \(r = 5\) and \(\frac { \mathrm { d } r } { \mathrm {~d} t } = 2\).

1. Show that \(r\) satisfies the differential equation $$\frac { \mathrm { d } r } { \mathrm {~d} t } = 0.08 r ^ { 2 }$$ [The surface area \(A\) and volume \(V\) of a sphere of radius \(r\) are given by the formulae \(A = 4 \pi r ^ { 2 }\), \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\).]
2. Solve this differential equation, obtaining an expression for \(r\) in terms of \(t\).
3. Deduce from your answer to part (ii) the set of values that \(t\) can take, according to this model.

10 \includegraphics[max width=\textwidth, alt={}, center]{dd7b2aee-4318-48e8-97c0-541e47f2e83a-4_335_875_262_635} A tank containing water is in the form of a cone with vertex \(C\). The axis is vertical and the semivertical angle is \(60 ^ { \circ }\), as shown in the diagram. At time \(t = 0\), the tank is full and the depth of water is \(H\). At this instant, a tap at \(C\) is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to \(\sqrt { } h\), where \(h\) is the depth of water at time \(t\). The tank becomes empty when \(t = 60\).

1. Show that \(h\) and \(t\) satisfy a differential equation of the form $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - A h ^ { - \frac { 3 } { 2 } } ,$$ where \(A\) is a positive constant.
2. Solve the differential equation given in part (i) and obtain an expression for \(t\) in terms of \(h\) and \(H\).
3. Find the time at which the depth reaches \(\frac { 1 } { 2 } H\).
   [0pt] [The volume \(V\) of a cone of vertical height \(h\) and base radius \(r\) is given by \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\).]

5 \includegraphics[max width=\textwidth, alt={}, center]{ccadf73b-16f5-463a-8f69-1394839d5325-2_346_437_1155_854} The diagram shows a variable point \(P\) with coordinates \(( x , y )\) and the point \(N\) which is the foot of the perpendicular from \(P\) to the \(x\)-axis. \(P\) moves on a curve such that, for all \(x \geqslant 0\), the gradient of the curve is equal in value to the area of the triangle \(O P N\), where \(O\) is the origin.

1. State a differential equation satisfied by \(x\) and \(y\). The point with coordinates \(( 0,2 )\) lies on the curve.
2. Solve the differential equation to obtain the equation of the curve, expressing \(y\) in terms of \(x\).
3. Sketch the curve.

4 The number of insects in a population \(t\) weeks after the start of observations is denoted by \(N\). The population is decreasing at a rate proportional to \(N \mathrm { e } ^ { - 0.02 t }\). The variables \(N\) and \(t\) are treated as continuous, and it is given that when \(t = 0 , N = 1000\) and \(\frac { \mathrm { d } N } { \mathrm {~d} t } = - 10\).

1. Show that \(N\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } N } { \mathrm {~d} t } = - 0.01 \mathrm { e } ^ { - 0.02 t } N .$$
2. Solve the differential equation and find the value of \(t\) when \(N = 800\).
3. State what happens to the value of \(N\) as \(t\) becomes large.

8 A certain curve is such that its gradient at a point \(( x , y )\) is proportional to \(\frac { y } { x \sqrt { x } }\). The curve passes through the points with coordinates \(( 1,1 )\) and \(( 4 , \mathrm { e } )\).

1. By setting up and solving a differential equation, find the equation of the curve, expressing \(y\) in terms of \(x\).
2. Describe what happens to \(y\) as \(x\) tends to infinity.

10 A gardener is filling an ornamental pool with water, using a hose that delivers 30 litres of water per minute. Initially the pool is empty. At time \(t\) minutes after filling begins the volume of water in the pool is \(V\) litres. The pool has a small leak and loses water at a rate of \(0.01 V\) litres per minute. The differential equation satisfied by \(V\) and \(t\) is of the form \(\frac { \mathrm { d } V } { \mathrm {~d} t } = a - b V\).

1. Write down the values of the constants \(a\) and \(b\).
2. Solve the differential equation and find the value of \(t\) when \(V = 1000\).
3. Obtain an expression for \(V\) in terms of \(t\) and hence state what happens to \(V\) as \(t\) becomes large.

10 A water tank is in the shape of a cuboid with base area \(40000 \mathrm {~cm} ^ { 2 }\). At time \(t\) minutes the depth of water in the tank is \(h \mathrm {~cm}\). Water is pumped into the tank at a rate of \(50000 \mathrm {~cm} ^ { 3 }\) per minute. Water is leaking out of the tank through a hole in the bottom at a rate of \(600 \mathrm {~cm} ^ { 3 }\) per minute.

1. Show that \(200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 250 - 3 h\).
   
   \includegraphics[max width=\textwidth, alt={}, center]{6280ab81-0bdb-47b4-8651-bff1261a0adf-17_2723_33_99_22}
2. It is given that when \(t = 0 , h = 50\). Find the time taken for the depth of water in the tank to reach 80 cm . Give your answer correct to 2 significant figures.

9. (i) Find $$\int \frac { ( 3 x + 2 ) ^ { 2 } } { 4 \sqrt { x } } \mathrm {~d} x \quad x > 0$$ giving your answer in simplest form.
(ii) A curve \(C\) has equation \(y = \mathrm { f } ( x )\). Given

• \(\mathrm { f } ^ { \prime } ( x ) = x ^ { 2 } + a x + b\) where \(a\) and \(b\) are constants
• the \(y\) intercept of \(C\) is - 8
• the point \(P ( 3 , - 2 )\) lies on \(C\)
• the gradient of \(C\) at \(P\) is 2
  find, in simplest form, \(\mathrm { f } ( x )\).
  

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15. \begin{figure}[h] \end{figure} Figure 4 shows a solid wooden block. The block is a right prism with length \(h \mathrm {~cm}\). The cross-section of the block is a semi-circle with radius \(r \mathrm {~cm}\). The total surface area of the block, including the curved surface, the two semi-circular ends and the rectangular base, is \(200 \mathrm {~cm} ^ { 2 }\)

1. Show that the volume \(V \mathrm {~cm} ^ { 3 }\) of the block is given by $$V = \frac { \pi r \left( 200 - \pi r ^ { 2 } \right) } { 4 + 2 \pi }$$
2. Use calculus to find the maximum value of \(V\). Give your answer to the nearest \(\mathrm { cm } ^ { 3 }\).
3. Justify, by further differentiation, that the value of \(V\) that you have found is a maximum.

13. \begin{figure}[h] \end{figure} Figure 3 shows a flowerbed. Its shape is a quarter of a circle of radius \(x\) metres with two equal rectangles attached to it along its radii. Each rectangle has length equal to \(x\) metres and width equal to \(y\) metres. Given that the area of the flowerbed is \(4 \mathrm {~m} ^ { 2 }\),

1. show that $$y = \frac { 16 - \pi x ^ { 2 } } { 8 x }$$
2. Hence show that the perimeter \(P\) metres of the flowerbed is given by the equation $$P = \frac { 8 } { x } + 2 x$$
3. Use calculus to find the minimum value of \(P\).

8. \begin{figure}[h] \end{figure} Figure 3 shows a flowerbed. Its shape is a quarter of a circle of radius \(x\) metres with two equal rectangles attached to it along its radii. Each rectangle has length equal to \(x\) metres and width equal to \(y\) metres. Given that the area of the flowerbed is \(4 \mathrm {~m} ^ { 2 }\),

1. show that $$y = \frac { 16 - \pi x ^ { 2 } } { 8 x }$$
2. Hence show that the perimeter \(P\) metres of the flowerbed is given by the equation $$P = \frac { 8 } { x } + 2 x$$
3. Use calculus to find the minimum value of \(P\).
4. Find the width of each rectangle when the perimeter is a minimum. Give your answer to the nearest centimetre.

9. A solid glass cylinder, which is used in an expensive laser amplifier, has a volume of \(75 \pi \mathrm {~cm} ^ { 3 }\).
The cost of polishing the surface area of this glass cylinder is \(\pounds 2\) per \(\mathrm { cm } ^ { 2 }\) for the curved surface area and \(\pounds 3\) per \(\mathrm { cm } ^ { 2 }\) for the circular top and base areas. Given that the radius of the cylinder is \(r \mathrm {~cm}\),

1. show that the cost of the polishing, \(\pounds C\), is given by $$C = 6 \pi r ^ { 2 } + \frac { 300 \pi } { r }$$
2. Use calculus to find the minimum cost of the polishing, giving your answer to the nearest pound.
3. Justify that the answer that you have obtained in part (b) is a minimum.

9. \begin{figure}[h] \end{figure} Figure 4 shows a plan view of a sheep enclosure.
The enclosure \(A B C D E A\), as shown in Figure 4, consists of a rectangle \(B C D E\) joined to an equilateral triangle \(B F A\) and a sector \(F E A\) of a circle with radius \(x\) metres and centre \(F\). The points \(B , F\) and \(E\) lie on a straight line with \(F E = x\) metres and \(10 \leqslant x \leqslant 25\)

1. Find, in \(\mathrm { m } ^ { 2 }\), the exact area of the sector \(F E A\), giving your answer in terms of \(x\), in its simplest form. Given that \(B C = y\) metres, where \(y > 0\), and the area of the enclosure is \(1000 \mathrm {~m} ^ { 2 }\),
2. show that $$y = \frac { 500 } { x } - \frac { x } { 24 } ( 4 \pi + 3 \sqrt { 3 } )$$
3. Hence show that the perimeter \(P\) metres of the enclosure is given by $$P = \frac { 1000 } { x } + \frac { x } { 12 } ( 4 \pi + 36 - 3 \sqrt { 3 } )$$
4. Use calculus to find the minimum value of \(P\), giving your answer to the nearest metre.
5. Justify, by further differentiation, that the value of \(P\) you have found is a minimum.

8. Liquid is pouring into a large vertical circular cylinder at a constant rate of \(1600 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) and is leaking out of a hole in the base, at a rate proportional to the square root of the height of the liquid already in the cylinder. The area of the circular cross section of the cylinder is \(4000 \mathrm {~cm} ^ { 2 }\).

1. Show that at time \(t\) seconds, the height \(h \mathrm {~cm}\) of liquid in the cylinder satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - k \sqrt { } h \text {, where } k \text { is a positive constant. }$$ When \(h = 25\), water is leaking out of the hole at \(400 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
2. Show that \(k = 0.02\)
3. Separate the variables of the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - 0.02 \sqrt { } h$$ to show that the time taken to fill the cylinder from empty to a height of 100 cm is given by $$\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { } h } \mathrm {~d} h$$ Using the substitution \(h = ( 20 - x ) ^ { 2 }\), or otherwise,
4. find the exact value of \(\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { h } } \mathrm {~d} h\).
5. Hence find the time taken to fill the cylinder from empty to a height of 100 cm , giving your answer in minutes and seconds to the nearest second.

9. Bacteria are growing on the surface of a dish in a laboratory. The area of the dish, \(A \mathrm {~cm} ^ { 2 }\), covered by the bacteria, \(t\) days after the bacteria start to grow, is modelled by the differential equation $$\frac { \mathrm { d } A } { \mathrm {~d} t } = \frac { A ^ { \frac { 3 } { 2 } } } { 5 t ^ { 2 } } \quad t > 0$$ Given that \(A = 2.25\) when \(t = 3\)

1. show that $$A = \left( \frac { p t } { q t + r } \right) ^ { 2 }$$ where \(p , q\) and \(r\) are integers to be found. According to the model, there is a limit to the area that will be covered by the bacteria.
2. Find the value of this limit. \includegraphics[max width=\textwidth, alt={}, center]{79ac81c3-cd05-4f28-8840-3c8a6960e7b7-31_2255_50_314_34}
   

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9. \begin{figure}[h] \end{figure} Figure 4 shows a cylindrical tank that contains some water. The tank has an internal diameter of 8 m and an internal height of 4.2 m .
Water is flowing into the tank at a constant rate of \(( 0.6 \pi ) \mathrm { m } ^ { 3 }\) per minute. There is a tap at point \(T\) at the bottom of the tank. At time \(t\) minutes after the tap has been opened,

• the depth of the water is \(h\) metres
• the water is leaving the tank at a rate of \(( 0.15 \pi h ) \mathrm { m } ^ { 3 }\) per minute
  1. Show that

$$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { 12 - 3 h } { 320 }$$ Given that the depth of the water in the tank is 0.5 m when the tap is opened,

find the time taken for the depth of water in the tank to reach 3.5 m .

4 Fig. 4 shows a cone. The angle between the axis and the slant edge is \(30 ^ { \circ }\). Water is poured into the cone at a constant rate of \(2 \mathrm {~cm} ^ { 3 }\) per second. At time \(t\) seconds, the radius of the water surface is \(r \mathrm {~cm}\) and the volume of water in the cone is \(V \mathrm {~cm} ^ { 3 }\). \begin{figure}[h] \end{figure}

1. Write down the value of \(\frac { \mathrm { d } V } { \mathrm {~d} t }\).
2. Show that \(V = \frac { \sqrt { 3 } } { 3 } \pi r ^ { 3 }\), and find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\).
   [0pt] [You may assume that the volume of a cone of height \(h\) and radius \(r\) is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\).]
3. Use the results of parts (i) and (ii) to find the value of \(\frac { \mathrm { d } r } { \mathrm {~d} t }\) when \(r = 2\).